Transcript Ch 7

Review – Last Week
 Sampling error
The radio station claims that on average a
household in San Diego spends $18 on candy
this Halloween.
A sample of 10 households reported that their
expenditure on candy is as follows:
$10
$20
$15
$30
$3
$20
$0
$15
$20
$7
What is the sampling error?
What do you think based on this sample?
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Mean
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More review
 Assuming the radio station also
reports that the variance of the
expenditure on candy is 90.
 Assuming the report is true, what
is the probability that your sample
mean is 14 or lower?
 What is the potential problem of
the probability evaluation method
you’ve used?
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Mean
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Chapter 7 Estimating
Population Mean
 In this chapter, we study how to use sampling
result to estimate population mean.
 Determine a confidence interval
• When population variance is known
• When population variance is unknown
 Determine the sample size to control the
estimation error
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Mean
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Point and Interval
Estimates
 Point estimate of population mean
 Use the sample mean – a single value.
Based on a result of clustered sampling, the
average housing price in San Diego county is
$495,000 in Sep 2005
Confident Interval
 Use a range to estimate.
Based on a result of clustered sampling, the
average housing price in San Diego county is
$495,000  $5,000 in Sep 2005.
What does this estimation mean?
What do you think about these two pieces of information?
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Mean
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Confidence Intervals
 Usually has the sample mean as
the middle point
 Is usually associated with a
confidence level.
 What is the probability that the
population mean is in the range
 In other words,
how confident are you?
 It provides more information about
a population characteristic than
does a point estimate
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Mean
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Determine confidence
intervals
 The estimation is normally a tradeoff
between confidence interval and
confidence level.
 The larger the interval, the higher the
confidence level. – but less useful
Based on a result of clustered sampling, the
average housing price in San Diego county is
$495,000  $400,000 in Sep 2005
-- with a confident level of 99%
 The smaller the interval, the lower the
confidence level. – less accurate
Based on a result of clustered sampling, the
average housing price in San Diego county is
$495,000  $50,000 in Sep 2005
-- with a confident level of 80%
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Mean
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Determine
the confidence level
In real life practice, the required
confidence level is normally given:
E.g. give your estimation about
the average annual income per
household in San Diego county
with 90% confidence level
The confidence level is always lower than 100%
Never 100% sure
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Mean
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Estimation process
Random Sample
Population
(mean, μ, is
unknown)
Mean
x = 50
Sample
I am 95%
confident that
μ is between
40 & 60.
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Mean
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When population  is
known
 Reminder:
x is normally distributed, with mean μ,
and standard deviaiton

n

95%
?
?
n
x
μx  μ
Most of the time,
you can get
good sample
z
x1
x2
But sometimes,
the sample is not
good. (unlucky)
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Mean
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When population  is
known
 Formula: the confidence interval
x  Z

2
n
where
-- (1   ) *100% is the confidence level;
-- Z  is the z-score which makes P(0<z<1)=
2
1-
;
2
-- x is the sample mean;
--

n
is the standard deviation of sample mean.
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Mean
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Steps to determine the
confidence interval
 Step 1: check whether sample mean x
is given
 If not, compute it.
 Step 2: check whether the standard
deviation of sample mean is given
 Sometimes, only population standard
deviation  is given. Divided it by n then.
 Step 3: use the required confidence
level to compute z/2
 Confidence level = 1-
 Probability = (1-)/2
 Check the reverse table for z/2
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Mean
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Example
 Problem 7.3 (a)
 Problem 7.4 (a)
(page 281)
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Mean
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When  is unknown
 We use the sample standard deviation
to estimate.
 How to calculate sample standard deviation
s ? (check chapter 3)
 Revised formula
x  t
2
s
n
-- s is the sample standard deviation
-- t/2 : the cutoff t-value from t-distribution
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Mean
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Student’s t-distribution
Standard Normal
(t with df = )
t (df = 13)
t (df = 5)
t
0
 A set of bell-shaped symmetric distributions
 Each has a degree of freedom: d.f.
 When df increases, the t-distribution gets closer
to normal distribution
 Formula for degree of freedom:
d.f. = n-1
 t-value for each x:
t 
x-μ
s
n
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Mean
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Get the cut-off t-value
from the table
 Use the table on page 597
“Values of t for selected probabilities”
 To check the table:
 First, get the degree of freedom.
e.g. d.f. =10
 And the confident level (e.g. 90%)
 When d.f. gets too large, use normal table
Conf. Level
0.7
0.8
0.9
0.95
0.98
0.99
One Tail
0.15
0.1
0.05
0.025
0.01
0.005
Two Tail
0.3
0.2
0.1
0.05
0.02
0.01
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Mean
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df
values of t
8
9
10
1.8125
11
12
Examples
 Problem 7.3 (b)
 Problem 7.4 (b)
 Problem 7.9 (P282)
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Mean
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Determine
the sample size
 When  is known:
 Confidence interval:
x  Z
2
 Z
2

n

n
is called the estimation error
 Sometimes, the estimation error is required not to
be too large
 Also, the confidence level (1-) is also required
 You have to get the large enough sample to
guarantee you meet both requirement.
Z
2
Z  


2
 required error  n  
 required error
n

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Mean




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2
Exercise
 Problem 7.25 (P. 288)
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Mean
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When  is unknown
 Need more complicated procedure
 Pilot sample: (Page 287)
 Start using a sample of n= 10 or 20.
 Get the sample mean and sample standard
deviation
 Use the sample standard deviation to
estimate the population standard deviation.
s
 Use
Z  

2
n 
 required error





2
to determine the sample size.
 Since we already have 20, n-20 more is still
needed.
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Mean
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Exercise
 Problem 7.27
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Mean
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